You seem to have a completely wrong idea about what's going on here!
No, c^X is not c^(1-2p). That would be cE(x) which is irrelevant.

What your probability function says is that x can have one of only two values: -1 and 1, with probabililty p and 1-p respectively. That means that c^x can have one of only two values: c-1= 1/c and c, with probability p and 1-p respectively. The "expected value" is the sum of the each possible value times its probability: p/c+ c(1-p)= 1. Solve that for c in terms of p.

Hey... I see the error I made now....
for solving... i managed to simplyfy the equation to 0 = (1-p)c^2 - c + p

This is the same as... 0 = (1-p)x^2 - x + p and the idea is to solve for x or p. Believe me, I tried EVERYYTHING to solve this, even the quadratic. I just keep finding my self at a ded end. I know that c=1 works just fine, but the question says that c cannot equal to one.
Anybody have any ideas?